Problem: Let $p$ and $q$ be the two distinct solutions to the equation $$(x-3)(x+3) = 21x - 63.$$If $p > q$, what is the value of $p - q$?
Solution: Factoring the right side, we can see that $(x-3)(x+3) = 21(x-3)$. Simplifying, we have $(x-3)(x+3) - 21(x-3) = (x-3)(x-18),$ and therefore our $p$ and $q$ are 18 and 3. Therefore, $p - q = \boxed{15}.$